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Six-fold Reality

Ian Beardsley

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Introduction……………………………………………………………………..3

The Theory……………………………………………………………………….4

Conclusion……………………………………………………………………….29

Continuing On………………………………………………………………….32

The Cosmic Calendar…………………………………………………………43

Appendix 1………………………………………………………………………..46

Appendix 2……………………………………………………………………….50

Appendix 3……………………………………………………………………….55

Appendix 4……………………………………………………………………….60

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Introduction

Here I show that the physical characteristics of the Universe can be described by the six-fold

unfolding of space and time in relatively simple terms, namely the characteristics inertia, proton

radius, proton charge, and the Sun’s magnetic field. Further that to do this we need to formulate

a constant that bridges the macrocosmos to the microcosmos, that is the planets to the proton

and the elements. Interestingly, their seems to be correspondence of this with the Old

Testament, which is discussed in terms of the cause being perhaps in the Universality of the six-

fold. The six-fold theory of reality results in a cosmic calendar, and it is correlated the Earth’s

past extinction events. It predict we could be in one now plus or minus 20 million years.

However many scientists believe we are in one now (Called the Anthropocene) where human

impact on the Earth is changing it drastically, mostly due to burning fossil fuels and putting CO2

into the atmosphere in excess of 350 parts per million, which causes global warming. We can

change that future, however, by switching to cleaner, renewable energy sources. We just have to

choose it.

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The Theory Years ago I saw a t-shirt that said:

God said Let there be light

And there was light.

It is one of Maxwell’s Equations, the one explaining that light (where c is the speed of light) is an

electromagnetic wave.

I was working on a theory for matter, space, and time, where matter, or inertia, was a property

of space and time, as well as the gravitational and electric fields, and that this was based in six-

fold symmetry. I sought not just to describe the microcosmos, the small atoms and fundamental

particles from which matter is made, but to describe the macrocosmos, the planets, Sun, and

moon, in terms of one another through some constant k that connected these two realms which

would be represented on the microscale by Planck’s constant h, and on the macroscale by the

universal constant of gravitation G. I had success with this and the work can be found in my

paper and book The Six-fold Nature of Reality, though we go into much of it here.

While working on this I found something very interesting. I was considering the Moon and its

kinetic energy and the Earth and its kinetic and six rotations of the Earth for my

theory of six-fold symmetry, and found that

Equation 1.

Equation 2.

Where

Equation 3.

Is a constant of proportionality and is the surface area of a proton, and is the rest energy

of a proton. is the fine structure constant. I found this interesting because aside from the

Bible’s Let there be light, and there was light, the light speed of light c squared is divided into 6

days, and we all know the Bible says God created the heaven’s and the Earth in six days, and

rested on the seventh. That is divided into 6 days it is the kinetic energy of the Moon to the

kinetic energy of the Earth by a factor of the fundamental particle from which matter is built, the

proton, its surface area and rest energy. The equation is approximate but it is closer to six days

than it is to five days or seven days. It is most accurate if you compute the kinetic energies of the

∇ × 𝔹 =

(

4π𝕁 +

∂𝔼

∂t

)

K E

4π r

K E

≈

6d a ys

κ ⋅

Pr otonSur fac Area

Pr oton RestEnerg y

⋅

Ear thKin et icE nerg y

Moon KineticEnerg y

6d a ys

LightSpeedSqu ared

κ =

4π r

of 5 62

Moon and the Earth using lunar orbital velocity at aphelion and the Earth orbital velocity at

perihelion (93% accurate). The six days comes from rounding to the nearest integer (6) because

the theory in which I was developing the idea was based on nearest integers as is done in

chemistry where you cannot have fractional neutrons when determining the number of neutrons

an element has from the molar mass and atomic number (See Appendix 1).

The equation comes from my equation for proton-seconds that relates the components of the

molecular skeletons of life hydrogen and carbon that make up the hydrocarbons, to time, and

which ultimately predicts the radius of a proton. The equations are:

Equation 4.

That is 1 second gives carbon. We find six seconds gives 1 proton is hydrogen:

Equation 5.

For time t greater than 6 seconds we have fractional protons. For t<1 we have the elements

heavier than carbon. 6 seconds is hydrogen, one second is carbon. I stumbled upon our

equations 1 and 2 which are what this paper is highlighting, when trying to eliminate time, t, in

equations 4 and 5. Indeed we see 6 seconds so closely gives 1 proton (hydrogen) and 1 second so

closely gives 6 protons (carbon) that we are compelled to look at the possibility that the duration

of a second is a natural unit that we accidentally created. The unit of a second came about from

dividing the day into 24 hours and that into 60 minutes and that into 60 seconds. We divided

the hour into 60 minutes and the minute into 60 seconds because our ancestors, the Sumerians,

and Babylonians, and from them the Ancient Greeks, chose to divide time by 60 because the 60

is so divisible evenly (evenly divisible by 1, 2, 3, 4, 5, 6, 8, 10, 12, 15 30,…). So that is something

natural that could be making the unit one second so functional in equations 4 and 5. Since there

are further about 30 days in a month, and 365.25 days close to the 360 degrees in a circle giving

the earth moving through about one degree a day, I made a first guess at the second as natural

by considering the the second as given by the Earth day by a factor of the kinetic energy of the

Moon and the the kinetic energy of Earth, which panned out nicely, I wrote:

Equation 6.

It is actually equal to 1.2 seconds. But this was not good enough because I sought a theory for

inertia the property that defines mass, from those constants which measure the properties of

space and time at both the microlevel (which would be Planck’s constant h) and and at the

macrolevel (which would be the universal constant of gravitation G). I stumbled upon a very

good lead in the work of Warren Giordano, which panned out nicely.

h 4π r

= 6pr oton ⋅ secon ds = carbon(C )

h 4π r

= 1pr oton ⋅ 6secon d s = hydr ogen(H )

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

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He wrote in his paper The Fine Structure Constant And The Gravitational Constant: Keys To

The Substance Of The Fabric Of Space, March 21, 2019:

In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic

equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where

‘ ’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational

constant numerically, but neglecting any units.

I found I could eliminated the and at the same time get the six of the six-fold symmetry

with which I was working by considering Avogadro’s number .

I suggested there exists some k that serves as a constant that describes both the microcosmos

and macrocosmos from the proton, to the atoms, to planetary orbits. It is such that the square

root of it times the earth orbital velocity is 6, because we are guessing we are dealing with six-

fold symmetry as the basis of Nature. That is

Eq. 7

We have that k is

Eq. 8

This follows from what Warren Giordano noticed that

Eq. 9

Without the right units. I noticed since Avogadro’s number is that I

could introduce an equation of state for the periodic tables of the elements:

Eq. 10

Eq. 11

Let us say we were to consider Any Element say carbon . Then in general

We have

and

Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a

molar mass of 12. We have

12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus

1 + α

6.02E 23atom s

k v

= 6

k =

8π

(1 + α) ⋅ N

𝔼

h(1 + α) ⋅ 10

= G

6.02 × 10

≈ 6 × 10

ℍ = 1

gra m

atom

(1 + α)

H = 6.0003

⋅ s

𝔼

ℂ

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

ℂ =

6gra m s

6protons

6(6E 23pr oton s)

6gra m s

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And it follows that

We see in general since the atomic number Z is the number of protons in an atom that in general

this holds for all elements because

And,

Therefore we always have:

is a variable, the number of protons in multiplied by Avogadro’s number.

Put in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:

Equation. 12

While we have masses characteristic of the microcosmos like protons, and masses characteristic

of the macrocosmos, like the upper limit for a star to become a white dwarf after she novas (The

Chandrasekhar limit) which is 1.44 — More mass than that and she will collapse — we do not

have a characteristic mass of the intermediary world where we exist, a truck weighs several tons

and tennis ball maybe around a hundred grams. To find that mass let us take the geometric

mean between the mass of a proton and the mass of 1.44 solar masses. We could take the

average, or the harmonic mean, but the geometric mean is the squaring of the proportions, it is

the side of a square with the area equal to the area of the rectangle with these proportions as its

sides. We have:

Equation. 13

We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .

We have the intermediary mass is:

Equation 14

All we really need to do now is divide equation 12 by equation 14 and we get an even number

that is the six of our six-fold symmetry.

ℂ = 6E 23

(1 + α)

⋅ N

ℂ = 6.0003kg

⋅

𝔼

Z ⋅ 6E 23protons

Z ⋅ gr a m s

Z ⋅ 6E 23protons

Z ⋅ gr a m s

⋅ 𝔼 = 6E 23

𝔼

(1 + α)

⋅ N

𝔼 ⋅ v

= 422.787kg

⊙

= 1.98847E 30kg

= 1.67262E − 27kg

= (2.8634E 30)(1.67262E − 27) = 69.205kg

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Equation. 15

The six of our six-fold symmetry.

We have something very interesting here. We have

This is:

Eq. 16

Where k is a constant, given

Eq. 17

It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf

will not have its gravity overcome the electron degeneracy pressure and collapse. The non-

relativistic equation is:

Let us approximate 0.77 with 3/4. Since we have our constant

Equation. 18

Equation 19

Then

Equation 20

Since our constant k in terms the Chandrasekhar limit is

Equation 21

The fine structure constant squared is the ratio of the potential energy of an electron in the first

circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of

light squared, that is it represents the ground state. It is

(1 + α)

⋅ N

𝔼 ⋅ v

= 6.1092 ≈ 6

69.205kg

6kg

= 6

k v

= 6

k =

773.5

M ≤ 0.77

ℏ

= 1.41 ⊙

k =

1 + α

⋅ N

𝔼

= Mm

ℏ

1/2

ℏ = h /2π

k =

8π

(1 + α) ⋅ N

𝔼

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Since

We are suggesting the earth orbit is the ground state for our planetary system. We suggest it

holds for any planetary system because k as we will see is a natural constant that solves many

physical problems on many levels, not just planetary systems but atomic systems and the

particles that make them up.

Let us now recall equations 4 and 5

While we have considered them to be proton-seconds because they are a mass divided by the

mass of a proton, we can consider these two masses to cancel and say they are equal to 1 second

and six seconds respectively. We have that carbon, which is to evaluate them at one second, is

the radius of a proton:

Equation 22.

This gives the radius of a proton is:

Equation 23.

Where . The experimental value of the proton radius is 0.833fm+/-0.014fm

Radius of hydrogen atom

Equation 24

k v

= 6

h 4π r

= 6pr oton ⋅ secon d s = carbon(C )

h 4π r

= 1pr oton ⋅ 6secon ds = hydr ogen(H )

h 4π r

= 6secon d s

4πh

= 8.288587 × 10

−16

= 0.829f m

= 1secon d

= 1.2E − 10m

t =

= R

⋅

h 4π

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Remember our constant k equation 21 (I find we have to divide by two somewhere and I think

this is because we are looking at packing protons here so they are offset by half their radius from

one another):

Equation 25

And 1 second in terms of the atom is given by

Then the equation for the radius of a proton is from equation 24:

Equation 26

Let’s verify our equation:

Making the approximation 9/8~1 we can write equation 26 as (We suggest we have picked up

the fraction 9/8 by making several approximations):

Equation 27

Which gives

We form constants:

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12secon d s

3 2

⋅

Gπ

⋅

𝔼

= 1.12secon d s

2 ⋅

4π

⋅

𝔼

1.67E − 27

(6.626E − 34)(299,792, 459)

(6.674E − 11)4π

⋅

1.2E − 10

6.02E 23

= 0.93f m

2π

⋅

𝔼

= 8.26935E − 16m ≈ 0.827f m

(

)

(

)

(

)

ℍ

(

)

of 11 62

Equation 28

Equation 29

And we have the Equation:

Equation 30

We can say that Avogadro’s number is not an arbitrary number because it is such that there are

twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add

to this section the equation of the mass of a proton.

Equation 31

We have successfully predicted the radius of a proton with a value for t in terms of the natural

constants. Let us return to equation 1:

Which we wrote as equation 2:

Using rather the second as

Which suggests the Moon plays some mysterious role in Nature. Indeed there is something

mysterious surrounding the Moon. If in the planet that has life orbiting a star there is an

indication to its intelligence that there is a mystery before it, it is that for our star the Sun, the

moon perfectly eclipses it as seen from the Earth. This is because:

Which are approximately equal. As well we can look at it as:

k =

2π

= 6.93E − 9kg

𝔼

= 1.99E − 34m

= k

𝔼

2π

⋅

𝔼

4π r

K E

≈

6d a ys

κ ⋅

Pr otonSur fac Area

Pr oton RestEnerg y

⋅

Ear thKin et icE nerg y

Moon KineticEnerg y

6d a ys

LightSpeedSqu ared

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

(lun ar − orbit)

(ear th − orbit)

384,400k m

149,592,870k m

= 0.00257

(lun ar − ra diu s)

solar − r a dius

1,738.1

696,00

= 0.0025

of 12 62

Which are about the same as well. The interesting thing is that since our ratios are around

0.0025 and 0.0045, then…

I say this is interesting because this the ratio of the precious metal gold (Au) to that of silver (Ag)

by molar mass, these elements being used for religious and ceremonial jewelry since ancient

times, is the same 1.8:

The sun is gold in color, the moon is silver in color. Thus the mystery is in that the moon is 400

times smaller than the Sun, but 400 times further from the Sun than it is from Earth and the

Solar Radius is 1.8 times the lunar orbital radius.

Indeed equation 22, which was

Tells us one second is in exactness for a value of time approximating a second;

Equation 32.

Which is very close to exactly one second, whereas

Equation 33.

It is at this point that we list the values of our constants that we are using and that we will be

using:

(lun ar − ra diu s)

(lun ar − orbit)

(1,738.1)

(384,400)

= 0.00452

solar − r a dius

ear th − orbit

696,000

149,597,870

= 0.00465

0.0045

0.0025

= 1.8

196.97

107.87

= 1.8

h 4π r

= 6secon d s

6α

h 4π r

= 1.004996352secon d s

K E

(Ear th Da y) = 1.2secon ds

of 13 62

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

And we want to look at how we formulated proton-seconds and how they are explaining the

nature of matter in terms of space and time.

Gravity is a property of space measured by the universal constant of gravity, G:

Equation 34

Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons

and neutrons) we will suggest given by:

Equation 35

Which describes mass per meter over time, which is:

Equation 36

It must be adjusted by the fine structure constant . It is my guess the factor should be which

is 18,769.:

Equation 37

Because the fine structure constant squared is the ratio of the potential energy of an electron in

the first circular orbit to the energy given by the mass of an electron in the Bohr model times the

speed of light squared, that is it represents the ground state. It is

Equation 38

Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a

hypersphere. Thus we consider the surface area of a proton, :

Equation 39

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G: 6.67408 × 10

−11

c : 299,792, 459m /s

α : 1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

G = 6.674E − 11N

= 6.674E − 11

⋅ kg

kg ⋅ s

1.82E − 16

kg ⋅ s

(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s /m

= 4π r

= 8.72E − 30m

of 14 62

We take the square root to get meters:

Equation 40

We multiply that with the value we have in equation 37:

Equation 41

We find that the mass of a proton realizes if we divide this by six seconds:

Equation 42

That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:

Equation 43

Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the

skeletons of organic matter, the so-called hydrocarbons:

Equation 44

Equation 45

If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of

the elements in the periodic table occur for dividing by something less that 1-second.

suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As

such there has to be a normal vector holding the higher dimensional sphere in a lower

dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in

the plane there should be a force countering it proportional to the normal holding it in a lower

dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a

sphere. This counter force would be experienced as inertia. Since Planck’s constant h is a

measure of energy over time where space and time are concerned it must play a role. Of course

the radius of a proton plays a role since squared and multiplied by it is the surface area of our

proton embedded in space. The gravitational constant is force produced per kilogram over a

distance, thus it is a measure of how the surrounding space has an effect on the proton giving it

inertia. The speed of light c has to play a role because it is the velocity at which events are

separated through time. The mass of a proton has to play a role because it is a measurement of

inertia itself. And alas the fine structure constant describes the degree to which these factors

have an effect. We see the inertia then in equation 6 is six protons over 1 second, by dimensional

analysis.

= 2.953E − 15m

(2.953E − 15m)(3.416E − 12

kg ⋅ s

) = 1.009E − 26kg ⋅ s

1E − 26

6secon d s

= 1.67E − 27kg

1E − 26kg ⋅ s

1secon d

= 1E − 26kg = 6proton s = 6m

1E − 26kg ⋅ s

6secon d s

1E − 26kg ⋅ s

1secon d

4π

of 15 62

We can make a program that finds integer solutions of

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by

the mass of a proton. But these masses can be considered to cancel and leave pure number. We

make a program that looks for close to whole number solutions so we can create a table of values

for problem solving.

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

0.2436 protons 24.750000 seconds 0.243635 decpart

h 4π r

Fig. 1

of 16 62

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of

the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons

the backbones of biological chemistry. Here is the code for the program:

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792459,protons[100],r=0.833E-15;

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?):

");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

float decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

Why did I write equation 23

Using 18/3 instead of 6? It is to emphasize the dynamics of six-fold symmetry.

4πh

= 8.288587 × 10

−16

= 0.829f m

of 17 62

The periodic table is periodic over 18 groups, which means if we proceed by number of protons

in an element, and start over after every 18 elements, the elements group themselves by their

natural properties, which is at the basis of doing chemistry and requires an entire textbook on

chemistry to go into fully, but here we need only point out that

The periodic table is periodic over 18 groups. 2, 3 are the smallest primes, the lowest factors in

which we can factor numbers, we have for the dynamics of six-fold symmetry:

In light of this we ask why six days (six rotations) of the Earth in equation 1:

The bible begins with God said let there by light and there was light, and it there goes into God

created the heavens and the earth in six days and rested on the seventh. Did someone in that

Middle Eastern desert eat a hallucinogenic plant and have a vision wherein he saw the

connection of the moon and earth with light and the six days of creation, or did he choose 6 days

for creation because he knew six was dynamic. Indeed the Bible speaks of the ratio of the

circumference of a circle to its diameter being 3. It is more accurately 3.141 but three is a close

estimate and it is not trivial to go on calculating the digits after the decimal. In fact the Ancient

Greek method of calculating pi begins with the estimate that the perimeter of a regular hexagon

to its diameter is three. The regular hexagon has six-sides and the property that its radius is

equal to its sides, and this is the power of six-fold symmetry because if its radius is one, each

side is one, of which there are six, and twice the radius is its diameter. Thus with six sides of one

each, the perimeter is 6, and the diameter is two giving pi=6/2=3. This regular hexagon

inscribed in a circle approximates a circle in all of its components, so thus approximates pi. If

the Bible says that the world was created in six days, perhaps that was because the regular

hexagon has six sides each equal in length. The seven day week came from the Romans 2000

years ago, the Old Testament was written 3500 years ago in Babylonian captivity. Since the

Moon goes around the earth in about 27 days, then if you are going to divide that into four units,

the week, the week is going to be 7 days. Seven times 4 is the approximately 29 day month, and

2 ⋅ 3 = 6

3 ⋅ 3 = 9

2 ⋅ 9 = 18

3 ⋅ 6 = 18

κ ⋅

Pr otonSur fac Area

Pr oton RestEnerg y

⋅

Ear thKin et icE nerg y

Moon KineticEnerg y

6d a ys

LightSpeedSqu ared

of 18 62

since the Moon goes around the Earth about 12 times in the time the earth goes around the Sun

once, there are 12 months in a year.

Indeed we can say the orbital velocities of the planets are all six if we give them in their natural

units. The natural units are if their orbits are approximately circular, and we call the radii of

such circular orbits as one in each case, we have a circumference of . Since is

approximately 3, then . The orbital velocities in each case are

where .

Thus we want to know the universal gravitational constant in these units:

We can immediately put this to work. In order for the earth to stay in orbit its centripetal force

must equal the gravitational force. Its orbital velocity must be given by:

If G=40, M, the mass of the sun equal to 1, and the orbital radius R is 1, then

Recall equation 7:

Then in terms of the constant k, all of the planetary orbits are given by the orbital velocity of

Earth, . At this time we return to equations 32 and 33:

2π r

2π r = 2(3)(1) = 6

2π r /T = 6/1 = 6

T = Pla n et′ sYear = 1

G = 6.67408E − 11

kg ⋅ s

G = 6.67408E − 11

kg ⋅ s

⋅

3.3479E 33m

⋅

1.98847E 30kg

⊙

⋅

9.9477E14s

year

G = 39.433

⊙

⋅ year

G ≈ 40

⊙

⋅ year

⊙

= 40

year

v =

(40)(1)

≈ = 6

k v

= 6

of 19 62

For our equation Earth day needs to be shorter. A long time ago it was; the Earth loses energy to the

moon. The days become longer by 0.0067 hours per million years. Our Equation

Is actually 1.2 seconds.

We have

24-20=0.0067t

t=597 million years

This was when the earth went through a dramatic change and there was a big explosion of life

(The Cambrian). This is when the Moon predicted the second as exactly 1. The dinosaurs went

extinct 65 million years ago giving small mammals a chance to evolve paving the way for

humans.

24-x=0.0067t

x=23.5645 hours

We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic

calendar:

What is the next term?

20+3+0.57735+0.4714=24hours

Which bring us to today.

6α

h 4π r

= 1.004996352secon d s

K E

(Ear th Da y) = 1.2secon ds

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

24h ours

1.2

= 20h ours

3cos(0

∘

) +

cos(30

∘

) = dinosaur − ex t in ct ion =

20hrs + 3hrs +

hrs +

of 20 62

We have discussed inertia and the proton, but we still need to discuss electric charge and the

solar magnetic field.

I construct the electric field as such (Fig. 2): is one component of . Events in

are through time with components and where c is the speed of light. The

electrons and the protons and in the presence of one another cross into giving

themselves acceleration in the and directions.

We have

We suggest

Equation 46

ℝ

(x, y, z)

ℝ

(x, y, z)

−x



0 ct

0 0 ct

= (ct

− ct

)



··

x ∝ (ct

− ct

)



Fig. 2

of 21 62

The position of changes in the space of and has travelled to a place in

time, as well. Since ct=meters there is a g such that . Thus since

and that is . Thus g is frequency squared ( ) and we suggest it is

derived from the separation between the charges and . That there is some velocity v such

that . We call upon our equation for 1 second:

Letting v= from our equation for k

And we have

Equation 47

We get

ℝ

(x, y, z)

t (0,ct

, ct

)

··

x = g(ct

− ct

)



g ⋅ m =

g =

⋅

⋅ s =

f =

(

)

h 4π

= 1secon d

(

)

(c)

h 4π

⋅

(c)

h 4π

= k

⋅

k =

8π

(1 + α) ⋅ N

𝔼 =

773.5

= 773.5

m = m

(

⋅

)

h 4π r

= q

h 4π r

(6.626E − 34)(4π)(0.833E − 15)

(6.674E − 11)(299,792, 459)

= 5.37E − 31

q = (0.033)

36(18769

)

773.5

(5.37E − 31) = 9.1435E − 19C

of 22 62

Equation 48

We model the formation of the solar system from a slowly rotating gas cloud, a nebula of

gaseous molecules, that collapses into a flat disc with a protostar at its center. The star turns on

and blows lighter elements far away, like hydrogen and helium, from which form the gas giants,

like Jupiter and Saturn, and the heavier elements stay closer in, like iron and silicates, from

which form the terrestrial planets like Venus, Earth, and Mars form. There are basically three

factors that determine its structure, the inward gravity, the pressure gradient outward which

balances with the inward gravity, and the outward inertial forces from the planets’ orbits. The

flattened rotating disc is broken up into rings each that has a mass spread out over it from which

the planets form. We estimate the ring associated with the Earth, had in its lower limit 230 earth

masses spread over it for the Earth to form. We further estimate that the Venus ring had a mass

spread over it of 230 Venus masses for Venus to form, and the Mars ring similarly had 230 Mars

masses spread out over it for Mars to form. The asteroid belt had about 200 of it masses, and the

Jovian planets 5, 8, 15, and 20 masses of each respectively. For Mercury it requires a factor of

about 350 because it is mostly iron condensations with incomplete silicon condensations.

Plotting these logarithmically we get the exponent of r, the distance of a planet from the sun is

-1.5 so that the density distribution of the protoplanetary disc is:

Giving a mass

With pressure gradients playing the key role in the formation of solar system, less attention is

payed to the magnetic field of the Sun. However, in the older literature, one of the pioneer’s of

this aspect found something very interesting concerning it. He was Alfven (1942). At the time

people were suggesting instead of the solar system forming from a rotating nebula, rather the

sun came into existence not at the same time at the center of the disc, but rather passed through

clouds and captured material after already existing. He figured for the captured material its

inward component v, and density , at a distance r from the sun, had to conserve mass, which

required:

He figured as the velocities of the atoms got closer to the sun, were moving then faster, collisions

would increase, and so temperature would go up, ionizing the atoms and therefore ionized, the

magnetic field becomes important. He considered for simplicity the solar magnetic field was

generated by a dipole moment , a vector quantity, and that a particle moving in the plane of

that vector with mass m and charge q, would have all of both the gravitational and magnetic

forces in that plane, so the problem becomes two-dimensional and required only the and , of

polar coordinates. The differential equations of its motion would be:

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71pr oton s ≈ 6pr oton s

σ (r) = σ

−3/2

= 3300

M =

∫

2π

∫

σ (r)rdrdθ

= 4π r

ρv

of 23 62

Equation 49

And,

Equation 50

We can integrate equation 50 with the boundary condition that the angular momentum of the

particle is zero at large distances from the sun to get (Appendix 2):

Equation 51

And substitute it into equation equation 49 for to get

Equation 52

Which we can write

Equation 53

We then integrate this with respect to r with the boundary condition that at large r and get

Equation 54

He then notices there is another value for which . It is

Equation 55

This is interesting because it means the particle can never approach the Sun closer than this

value, and it depends only on the value q/m, the charge to mass ratio of a particle. He took this

as hydrogen because ionized it is a proton, for which q/m is well defined. He estimated what the

magnetic field of the Sun could have been in this earlier stage of its life, and adjusted for the fact

that hydrogen doesn’t ionize until it reaches a velocity of 5E4 m/s and found that was the

region occupied by the major planets which are Jupiter and Saturn mostly made of hydrogen

and helium.

Certainly today we don’t see the planets as having formed from material gathered by the Sun in

its journey, but rather think the Sun and planets formed at the same time from a cloud that

collapsed into a rotating flat disc. And indeed, there may be stars in the galaxy that pass through

clouds and gather material, and indeed Alfven’s equations would hold preventing ionizing

clouds from falling into their star. But we can also apply his equation to our Sun today, for which

we know a great deal about its magnetic field, which also happens to be an important thing to

study and for which we have satellites in the Lagrange points, where the Earth’s gravity cancels

··

r = −

⊙

qμ

+ mr

d(r

θ )

= −

qμ

θ = −

∫

qμdr

qμ

··

r = −

⊙

= −

⊙

r = 0

2GM

⊙

−

r = 0

(

2GM

⊙

)

1/3

of 24 62

with that of the Sun, where the orbits are very stable, so we can understand the solar magnetic

field. It is a complex field, that interacts with the Earth’s magnetosphere, and we need to predict

solar maximums, so we have warning as to whether there will be a magnetic storm that will

knock out our electrical grid and internet, ahead of time.

During solar minimum the solar magnetic field has closed lines, that flow out one pole and into

the other. The dipole field of the sun is about 50 Gauss. There are 10,000 Gauss in a Tesla, so

that is 5E-3 Tesla. That is the magnetic field strength where the field goes into the poles. The

total magnetic field of the Sun at the Earth, is all the components taken together, which are

, , and . The important component is , because it runs north-south, so it is

perpendicular to the ecliptic, the path traced out by the sun due to the earth’s orbit. It is the

component that interacts with the Earth magnetosphere, and when it points southward, it will

connect with the Earth’s magnetosphere which points northward so the solar poles flow into the

Earth poles and the Earth field then gets disrupted allowing particles from the solar wind to rain

down along Earth magnetic field lines causing the Aurora. The solar magnetic field doesn’t

always stay around the Sun itself, but the solar wind carries it through the solar system until it

collides with the interstellar medium reaching the heliopause. Thus the Sun creates the

Interplanetary Magnetic File (IMF) which has a spiral shape because the Sun rotates once about

every 25 days. But the upshot is that at Earth we have

Moderate Magnetic Field: 10 nT

Strong Magnetic Field: 20 nT

Very Strong Magnetic Field: 30 nT

For our purposes we want to return to equation 55:

of 25 62

And ask just what is , because in the time that Alfven was working we worked with magnetic

fields differently, aside from his equation uses a trick, which we still use today, and that is to

consider the magnetic field a dipole. To consider it like this is to say there are two monopoles

opposite in polarity. According to Maxwell’s equation we cannot have magnetic monopoles,

though they are predicted by some modern theories, they have never been found. The trick is in

that by treating the North magnetic pole and South magnetic pole as separate magnetic charges

is to treat them like we do electric charges, the charge of a proton and the charge of an electron,

which can be convenient for making computations, but don’t exist that we know of. So we will

solve equation 56 for , and see what its units are so we can understand what it represents and

we will let m be the mass of a proton and q the charge of a proton. We get:

Equation 56

We can write these units as, by taking Coulombs (C) equal to

Equation 57

This is units of force per current density, which makes sense because a flowing current creates a

force. We can also write it:

Equation 58.

Which is energy per magnetic field strength in that the SI units of magnetic field strength is

amps per meter. This tells us:

Equation 59

Thus we will use the energy as ionization of hydrogen, the energy to remove its electron and

make it a proton:

We have

, , ,

(

2GM

⊙

)

1/3

⊙

= μ =

⋅ kg

C ⋅ s

a mp ⋅ secon d s

⋅

a mps

⋅

a mps

μ =

Energ y

Magnet icFiel d St rength

H ⟶ H

+ e

−

= 1pr oton = 2.18E − 18J

= 1.6E − 19C

= 1.67E − 27kg

G = 6.67408E − 11N

⊙

= 1.989E 30kg

of 26 62

We want to look at the Sun as having a current flowing around its equator in a loop with its

radius

We find for the dipole field of the Sun at 50 Gauss=5E-3T, which is about 100 times stronger

than the Earth magnetic field, that this is a current I=5.5362E12 amperes driving the solar

magnetic dipole. This gives us that since the Earth orbit (1AU=1.495979m):

Equation 60

Equation 61

Or,…

Equation 62

The radius of a proton is 0.833E-15m. We have that r is:

Equation 63

⊙

= 6.957E 8m

1AU

5.5362E12amps

1.496E11m

= 37.0a m ps /m

μ =

Ioni zat ionE nerg y

Magnet icFiel d St rength

2.18E − 18J

37.0A mps /m

= 5.892E − 20

J ⋅ m

r =

(

2GM

⊙

)

1/3

= 4.9324E − 15m

= 5.92 ≈ 6Pr oton Ra dii = carbon

of 27 62

Our six-fold symmetry unfolding. This is again the carbon the core element of life. We see it

provided for by the Sun’s magnetic field.

of 28 62

Let us write the computation as one equation, and verify it. We have

Where

=Ionization energy of Hydrogen

=3.47E-39 (correct)

Equation 64

(0.000034574)(3.47E-39)=1.19972E-43

1.19972E-43^(1/3)

=4.932E-15

So as you can see equation 62 is correct. It says that carbon, the basis of life is in the ratio of the

solar magnetic field and the solar gravitational field.

r =

(

2GM

⊙

)

1/3

= 4.9324E − 15m

⊙

1AU

5.5362E12amps

1.496E11m

= 37.0a m ps /m

1AU = r

μ =

Ioni zat ionE nerg y

Magnet icFiel d St rength

2.18E − 18J

37.0A mps /m

= 5.892E − 20

J ⋅ m

⊙

1AU

⊙

= (5.982E − 20)

= 3.47E − 39

(IE

)

⊙

(2.18E − 18)

(12.56637E − 7)

(1.496E11)

4(5E − 3)

(6.957E8)

(

2GM

⊙

⋅

(IE

)

⊙

)

1/3

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(1.496E11m)

4(5E − 3T )

(6.957E8m)

1/3

of 29 62

Conclusion

Indeed we have shown that we can predict the radius of a proton and its mass in terms of the

properties of space:

Which gives

And the interesting thing is they do not yet have an equation for the radius of a proton, yet we

have it here and it doesn’t just use Planck’s constant, h, of the indeterministic microcosmos, but

uses the gravitational constant, G, of the deterministic macrocosmos. We have further shown

our theory of inertia is based on six-fold symmetry is centered around hydrocarbons, the

skeletons of life chemistry

We find 1 second gives

We find six seconds gives 1 proton is hydrogen:

We further find this same sixfold symmetry describes not just proton systems (atoms) but our

solar system

If is the orbital radius of the moon, and is its orbital period, , Venus and so on. We

also have that the Earth orbit is the ground state for the solar system:

Where,

And

2π

⋅

𝔼

= 8.26935E − 16m ≈ 0.827f m

h 4π r

= 6pr oton ⋅ secon d s = carbon(C )

h 4π r

= 1pr oton ⋅ 6secon ds = hydr ogen(H )

(

6 6 6

)

k v

= 6

k =

8π

(1 + α) ⋅ N

𝔼

of 30 62

We further find that for the moon

Where six days is six rotations of the Earth. This shows the earth/moon/sun system and the

proton are structured around one another in terms of six-fold symmetry. Since there are 360

degrees in a circle, and the earth year is close to 360 rotations of the earth (365.25) the earth

orbit can be divided into six sixty degree sectors, the earth moving through about one degree a

day in its orbit.

We have further found that the solar magnetic field follows through with the same six-fold

symmetry, accommodating carbon:

We also have that

Which is to say

Where

In our equation

(

⋅

)

h 4π r

= q

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71pr oton s ≈ 6pr oton s

h 4π r

⋅

= 6d ays

(

2GM

⊙

⋅

(IE

)

⊙

)

1/3

4π r

K E

≈

6d a ys

κ ⋅

Pr oton Sur fac Area

Pr oton RestEnerg y

⋅

Ear thKinet icEnerg y

Moon KineticEnerg y

6d a ys

LightSpeedSqu ared

κ =

h 4π r

= 6pr oton ⋅ secon d s = carbon(C )

of 31 62

In terms of the Moon the radius of the proton is:

Where . The experimental value of the proton radius is 0.833fm+/-0.014fm. We

can also for a constant k, different than our former constant for which we can write

And we have the Equation:

In our equation

Anything less than a second is heavier than carbon, the core element of life, and the unit of a

second in this is given by the moon, as we said

This brings us to ask is there a reason the Moon is important to life? The answer is yes, the Earth

is tilted to its orbit by 23.5 which results in the seasons and the Moon in its orbit holds the Earth

at this angle, if it didn’t we would have wild, extreme seasons. If the tilt was less we might not

have season at all, and more the seasons might be too extreme.

K E

moon

K E

earth

(Ear th Da y) ≈ 1second

4πh

= 8.288587 × 10

−16

= 0.829f m

= 1secon d

k =

2π

= 6.93E − 9kg

𝔼

= 1.99E − 34m

= k

𝔼

h 4π r

= 6pr oton ⋅ secon d s = carbon(C )

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

of 32 62

Continuing On At this point we suggest the reason the Sun is in balance between its

gravitational and magnetic fields at Earth orbit for 6 proton radii is that it is to accommodate for

carbon’s central life functions. The radius of a carbon nucleus is 2.7E-15m not to be confused

with its radius, which is the distance from its center to its first electron orbital, which is

7.05E-11m or 26,000 times larger. Thus we consider equation 62

And see how many times it fits into the radius of the nucleus of carbon:

We then notice that this is the ratio of the molar mass silver (Ag) to that of gold (Au) which are

the finest electrical conductors, silver the finest at room temperature, gold the finest at extreme

temperatures, to get:

Which is exact to two places after the decimal 0.55. We take this moment to recall that the lunar

orbital radius to the solar radius is in the same as silver to gold as well, and that the Moon is

silver in color and the Sun is gold in color and that silver and gold have been used for ceremonial

jewelry since ancient times, and further recall equation 2:

We mention these things but mostly want to point out that the Moon has a function that serves

the viability of life on Earth. It is that the Earth is tilted to its orbit by 23.5 which results in the

seasons and the Moon in its orbit holds the Earth at this angle, and if it didn’t we would have

wild, extreme seasons. If the tilt was less we might not have seasons at all, and more the seasons

might be too extreme.

A picture here starts to form suggesting that life on Earth has components in the solar magnetic

field and the moon as connected with one another through electrical conductors such as the

finest gold and silver.

While gold and silver are not magnetic in any practical sense, being conductors of course with a

current run through them they can create magnetic fields, but interestingly the silicon mantle

over the core has immense amounts of gold trapped in it, enough to cover the Earth’s surface 13

inches high. It is at about 1,800 feet deep and at temperatures of many thousands of degrees.

The core is also enriched with gold and silver where temperatures are 5,400 degrees C and

pressures are up to 360 gigapascal. Actually there is enough gold in the Earth’s core to coat its

surface a foot and a half deep. The magnetic field of the earth is mostly caused by molten iron

and nickel in the Earth’s outer core; Their convective kinetic energy is converted to electric and

magnetic energy. The convection is caused by heat escaping from the core, which is a natural

process called a geodynamo.While the solar magnetic field alternates between minimum and

maximum, and flips, we may be arriving at a way to define it precisely in terms of carbon.

r =

(

2GM

⊙

)

1/3

= 4.9324E − 15m

2.7E − 15

4.932E − 15

= 0.5474

196.97

107.87

= 0.5476

κ ⋅

Pr otonSur fac Area

Pr oton RestEnerg y

⋅

Ear thKin et icE nerg y

Moon KineticEnerg y

6d a ys

LightSpeedSqu ared

of 33 62

We found the value for where the solar magnetic field balances with solar gravity results in

carbon occurring at Earth orbit, and carbon is the core element of life which is very abundant on

Earth. I have computed the values for the rest of the planets, and they don’t land on integer

values that line up with the integer values of protons in the elements. However, I did this for

Jupiter’s magnetic field in balance with its gravity for all of the Galilean satellites and they all

line-up with integer values closely. We have

Io: 14 protons. Is silicon (Si) with molar mass 28.09 grams/mole

Europa: 19 protons Is potassium (K) with molar mass 39.10 grams/mole

Ganymede: 26 protons Is iron (Fe) with molar mass 55.85 grams/mole

Callisto: 38 protons Is strontium (Sr) with molar mass 87.62 grams/mole

We see if we start with the period 2 inert gas neon (20 grams/mole) period 2 being the period

that has the most abundant life elements C, N, O and neon being an inert gas (group 18, the last

group) it determines the properties of all the elements in period 2, and the first three Galilean

satellites are produced by loop logic in computer programming. That is:

(7/5)(Neon)=(7/5)(20)=28 g/mol = Silicon = Io

We put that value in our equation (This is the loop logic):

(7/5)(Silicon)=(7/5)(28)=39.2 g/mol= Potassium = Europa

Put the result in our equation again:

(7/5)(Potassium)=(7/5)(39.2)= 54.88 = Iron =Ganymede

And, again…

(7/5)(Iron)=(7/5)(54.88)= 76.832 Is not equal to Callisto

We can write a few lines of code (Here in C) that demonstrates this:

#include <stdio.h>

int main(int argc, const char * argv[]) {

float Ne=20.00;

while (Ne<77)

{

Ne=(1.4)*Ne;

printf("%f \n", Ne);

}

return 0;

}

of 34 62

Running The Program

28.000000

39.200001

54.880001

76.832001

107.564804

Program ended with exit code: 0

All we really need is an equation for the distribution of the Galilean satellites, so we can turn the

above into an equation. And, this we can do quite nicely:

Io: 4.218E8m = 1

Europa: 6.71E8m = 1.591

Ganymede: 1.0704E9m = 2.5377

Callisto: 1.8827E9m = 4.4635

And, the equation is:

We now go on to present all of computations for all of the planets and the Galilean satellites.

= 1.6

= 1

= 1.6

= 2.56

= 4.096

of 35 62

Europa Orbiting Jupiter

Jupiter Magnetic Field: North 14 Gauss, South 11 Gauss

(14+11)/2=12.5 Gauss = 1.25E-3 Tesla = 12.5/0.5 = 25 Times Earth magnetic field

Jupiter mass: 1.898E27 kg

Jupiter radius: 6.9911E7 m

Europa orbits at: 6.71E8 meters

= 1.58842E-14m

1.58842E-14/0.833E-15=19.06867 proton radii = Potassium (K) almost exact 39.10 g/mol

Io orbiting Jupiter

=1.1656E-14/0.833E-15=13.993~14 proton radii = Silicon (Si) almost exact 28.09 g/mol

Ganymede Orbiting Jupiter

=2.1686E-14/0.833E-15=26.0337 proton radii~26 protons = Iron (Fe) almost exact 55.85 g/mol

Callisto Orbiting Jupiter

=3.15988E-14/0.833E-15=37.933 proton radii~38 protons = Strontium (Sr) almost exact

87.62 g/mol

19r

(1.6E − 19C )

2(6.67408E − 11N

)(1.898E 27kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(6.71E8m)

4(1.25E − 3T )

(6.9911E 7m)

1/3

14r

(1.6E − 19C )

2(6.67408E − 11N

)(1.898E 27kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(4.218E8m)

4(1.25E − 3T )

(6.9911E 7m)

1/3

26r

(1.6E − 19C )

2(6.67408E − 11N

)(1.898E 27kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(1.0704E 9m)

4(1.25E − 3T )

(6.9911E 7m)

1/3

57r

(1.6E − 19C )

2(6.67408E − 11N

)(1.898E 27kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(1.8827E 9m)

4(1.25E − 3T )

(6.9911E 7m)

1/3

of 36 62

Magnetic Field at Solar Radius

1.75E-22 m

1.75E-22/0.833E-15 = 0.000000210 proton radii

Mercury Orbit

46,00,000 to 70,000,000 kilometers

(46+70)/2=58,000,000,000 meters

=1.340E-18m

1.340E-18/0.833E-15=0.0016086 proton radii = 1 quark

Venus Orbit

=3.97347E-15m

3.97347E-15/0.833E-15=4.770 proton radii~boron = 5 protons = Boron (B)

Earth Orbit

(0.000034574)(3.47E-39)=1.19972E-43

1.19972E-43^(1/3)

=4.932E-15

4.932E-15/0.833E-15=5.92 proton radii=6 protons= Carbon (C)

0.000000210r

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

4(5E − 3T )

1/3

0.0016r

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(5.8E10m)

4(5E − 3T )

(6.957E8m)

1/3

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(1.082E11m)

4(5E − 3T )

(6.957E8m)

1/3

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(1.496E11m)

4(5E − 3T )

(6.957E8m)

1/3

of 37 62

Mars Orbit

=6.532E-15m

6.532E-15/0.833E-15=7.84 proton radii = 8 protons = Oxygen (O)

Asteroids

2.2-3.2 AU=2.2(1.496E11m)-3.2(1.496E11m)=3.29E11m-4.787E11m

(1.496E11+4.787E11)/2=3.1415E11 meters

=8.08794E-15m

8.08794E-15/0.833E-15=9.7094 proton radii=10 protons = Neon (Ne)

Jupiter Orbit

1.48E-14m

1.48E-14/0.833E-15=17.767 proton radii=protons= Argon (Ar)

Saturn Orbit

=2.2183E-14m/0.833E-15m=26.6308=27 protons=cobalt (Co)~Aluminum=26.98 g/mol

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(2.28E11m)

4(5E − 3T )

(6.957E8m)

1/3

10r

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(3.1415E11m)

4(5E − 3T )

(6.957E8m)

1/3

18r

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(7.78E11m)

4(5E − 3T )

(6.957E8m)

1/3

27r

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(1.427E12m)

4(5E − 3T )

(6.957E8m)

1/3

ℍ = 1

gra m

atom

𝔼 =

Z ⋅ gr a m s

Z ⋅ protons

Z ⋅ 6E 23protons

Z ⋅ gr a m s

⋅ 𝔼 = 6E 23

of 38 62

Uranus Orbit

=3.540E-14/0.833E-15=42.50 protons is between Tellurium and Iodine.

=4.77E-14/0.833E-15=57.2677 proton raddi=Lanthanum (La)

Mean Pluto Orbit

5.71454E-14/0.833E-15=68.60 proton radii=erbium (Er) which makes it outside the main block

of the periodic table in the Lanthanide series.

43r

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(2.877E12m)

4(5E − 3T )

(6.957E8m)

1/3

57r

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(4.5E12m)

4(5E − 3T )

(6.957E8m)

1/3

68r

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(5.9E12m)

4(5E − 3T )

(6.957E8m)

1/3

of 39 62

There have been five mass extinctions, and it is suspected we are experiencing the sixth right

now. We identify the times when these happened by noticing the disappearance of species in the

geologic record. The five great mass extinctions where more than 75% of the species disappeared

were:

444 million years ago, the Ordovician. 86% of species lost.

The cause was a short, severe ice age. The cause is thought to be by perhaps the uplifting of the

Appalachians, exposing silicate rock which sucked the CO2 out of the atmosphere that kept the

planet warm.

375 million years ago, The Devonian. 75% of species lost.

The cause was probably the arrival of new land plants covering the planet whose deep roots

stirred up the earth releasing nutrients into the ocean which might have caused algae blooms

that sucked oxygen out of the water suffocating the life that dwelled at the ocean bottom (for

example trilobites).

251 million years ago, The Permian. 96% of species lost

This was the worst extinction and caused by a cataclysmic eruption near Siberia which blasted

CO2 in the atmosphere, causing methanogenic bacteria to belch methane, one of the most

powerful greenhouse gases causing temperatures to rise, oceans to acidify and stagnate and

release the poisonous hydrogen sulfide.

200 million years ago, the Triassic. 200 million years ago. 80% of the species lost. They have not

been able to find a cause of this extinction.

66 million years ago, The Cretaceous. 76% of species lost.

Volcanic activity and climate change was the cause for the ammonites in the ocean as well there

was with asteroid impact that brought the dinosaur’s end.

Cosmic Rays, their story begins in 1912 when the Austrian physicist Victor Hess took a balloon

5.3 km up to measure radiation levels in the atmosphere. He found radiation levels increased

with altitude and concluded radiation was coming in from space. This radiation came to be

called “Cosmic Rays”. They are charged subatomic particles. There are two theories for their

origins: They come from massive black holes at the center of galaxies or from gamma ray bursts

(GRB’s) caused by super massive stars in distant galaxies that nova and collapse into a neutron

star or black hole. When these stars run out of fuel their radiation pressure no longer counters

gravity and the star collapses into a blackhole, and energy may be released along the axis of

rotation of the star creating a gamma ray burst. Bursts can last 10 milliseconds to several hours.

While these bursts come from distant galaxies, which must have been younger than ours when

the star went nova because the further you look, the further back in time you are looking, a

similar, weaker burst can happen in our galaxy called a soft gamma repeater which is associated

with a kind of star called a magnetar; a kind of star recently hypothesized in 1992 by Robert

Duncan and Christopher Thompson that is a type of neutron star with an extremely powerful

magnetic field (10^9 to 10^15 Gauss) which powers x-rays and gamma rays by its decay. It has

been hypothesized if one caused a gamma ray burst that pointed directly towards the Earth, it

could cause a mass extinction event. Some researchers have hypothesized such a burst was

responsible for the Ordovician mass extinction.

Interestingly, the energy source for the prebiotic path to life (prebiotic synthesis) could have

been powered by cosmic rays, (protons of 2.5 to 3.0 MeV). According to Google “High energy

of 40 62

protons are a major component of cosmic rays, and there was likely a significant flux of cosmic

rays on the early Earth.” It is becoming more thought to have been that the building blocks of

life came from ice crystals in interstellar space, or from the same in meteorites and comets. Not

just the sugar of the DNA and RNA phosphate backbone, ribose, but the nucleobases as well,

where meteorites studied have been found to have three of the five adenine, guanine, and uracil

(2022). As well amino acids that are synthesized into proteins from DNA instructions.

Recently it has been discovered in the MICMOC experiment that started in 2003 at IAS that

through a non-directed method, as opposed to methods where results are achieved by synthesis

in the laboratory, which it must be to simulate something happening of its own accord in

conditions that were thought to exist when life got its start, they found the prebiotic formation of

ribose, where prebiotic means before life and leading up to it.

They have found from studying photochemistry and thermochemistry of laboratory ices

analogous to interstellar and cometary ices, prebiotic molecules, such as amino acids in the

organic residues from sublimation of the ices. Two very important residues were discovered,

glycolaldehyde of 2 carbon atoms and glyceraldehyde of 3 carbon atoms. The sugar ribose was

produced in a formose reaction that was natural and undirected, which was a first for ribose in

that the formose reaction, discovered by Aleksandr Butlerov, was not prebiotic but rather

directed in that it required a base catalyst, where with the comet ices they used the presence of

solvated electrons that are labile in the ice acted as a catalyst. It also produces much larger

quantities of ribose. The reaction is a polymerization of formaldehyde using glycolaldehyde and

glyceraldehyde.

Back to cosmic rays. DNA and RNA are twisted ladders, a spiral helix and this makes them

chiral, but they can rotate one way or the other, left or right, and life chose one giving it a

handedness. It has been suggested that because comic rays cause mutations and thus affect

evolution and since cosmic rays produce muons, which produce electrons and positrons, that

have their magnetic moments pointing in a direction opposite their velocity, making them

polarized and always with the same handedness, and since they can interact with biological

molecules, they can give the molecules over evolutionary time the choice of one handedness over

the other, through the weak interaction. Others argue the handedness comes from local

magnetic fields, or polarized light.

Still there are those who argue, like Trevors and Abel (2004) that

“Peer-reviewed life-origin literature presupposes that, given enough time, genetic instructions

arose via natural events. Thus far, no paper has provided a plausible mechanism for natural-

process algorithm-writing. Only 200 million years separated the of Earth’s bombardment from

the presumed first appearance of life on Earth 3.8 billion years ago.”

“The genetic operating system uses a bijective coding system whereby a certain triplet codon

represents a certain amino acid…How did inanimate nature write (1) the conceptual instructions

needed to organize metabolism (2) a language/operating system needed to symbolically

represent, record and replicate those instructions? (3) a bijective coding scheme (a one-to-one

correspondence of symbol meaning) with planned redundancy so as to reduce noise pollution

between triplet codon “block code” symbols (“bytes”) and amino acid symbols? (4) How did

inanimate nature design and engineer a cell (Turing machine?) capable of implementing those

coded instructions.”

In short, Trevors and Able are raising a point that many biologists are arguing, namely, not just

that DNA can evolve through mutations, but how did a mechanism get there that can evolve, a

complete language that can encode something as complex as life, be read, and copied to

reproduce.

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Further problems arise. Referring to Trevors and Abel describing the cell as a Turing machine,

ribosomes synthesize proteins by translating the code transcribed in mRNA into an amino acid

sequence. But they need to be synthesized themselves from amino acids, and where did the first

ribosome come from that did this?

Finally, all of that translating and describing and unzipping of DNA when it replicates, is a

process by which you would think it needs a mind to tell it what to do and control it. They have

looked for such a mind in the nucleus of cells and have not been able to find one. Further you

can say the brain does not do it because it has no nerves going from brain to the cell to the

ribosome to connect them. This is known as “The Mindless Cell Paradox.”

Stanley Miller in 1952 was working under Harold Urey, and came up with the idea of mixing

theoretical primordial elements and compounds in a bottle under electric energy to see if

biological elements would arise spontaneously. He hypothesized the early Earth atmosphere

would be composed of methane, ammonia, and hydrogen and since it had water vapor clouds

there would be lightning. The result was he produced 11 of the 20 biological amino acids. The

formose reaction is thought to parallel the reactions that lead to sugars like ribose. The Strecker

reaction is thought to parallel the reactions that produced amino acids on Earth or on meteorites

that delivered them to Earth, so-called carbonaceous chondrite meteorites.

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The Cosmic Calendar Let us return to our cosmic calendar:

For our equation Earth day needs to be shorter. A long time ago it was; the Earth loses energy to the

moon. The days become longer by 0.0067 hours per million years. Our Equation

Is actually 1.2 seconds.

We have

24-20=0.0067t

t=597 million years

This was when the earth went through a dramatic change and there was a big explosion of life

(The Cambrian). This is when the Moon predicted the second as exactly 1. The dinosaurs went

extinct 65 million years ago giving small mammals a chance to evolve paving the way for

humans.

24-x=0.0067t

x=23.5645 hours

We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic

calendar:

What is the next term?

20+3+0.57735+0.4714=24hours

Which bring us to today. Let us see if we can make it more complete by looking at all of the five

great extinctions, not just the Cretaceous of the dinosaurs:

6α

h 4π r

= 1.004996352secon d s

K E

(Ear th Da y) = 1.2secon ds

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

24h ours

1.2

= 20h ours

3cos(0

∘

) +

cos(30

∘

) = dinosaur − ex t in ct ion =

20hrs + 3hrs +

hrs +

of 44 62

444 million years ago, the Ordovician. 86% of species lost.

The cause was a short, severe ice age. The cause is thought to be by perhaps the uplifting of the

Appalachians, exposing silicate rock which sucked the CO2 out of the atmosphere that kept the

planet warm.

375 million years ago, The Devonian. 75% of species lost.

The cause was probably the arrival of new land plants covering the planet whose deep roots

stirred up the earth releasing nutrients into the ocean which might have caused algae blooms

that sucked oxygen out of the water suffocating the life that dwelled at the ocean bottom (for

example trilobites).

251 million years ago, The Permian. 96% of species lost

This was the worst extinction and caused by a cataclysmic eruption near Siberia which blasted

CO2 in the atmosphere, causing methanogenic bacteria to belch methane, one of the most

powerful greenhouse gases causing temperatures to rise, oceans to acidify and stagnate and

release the poisonous hydrogen sulfide.

200 million years ago, the Triassic. 200 million years ago. 80% of the species lost. They have not

been able to find a cause of this extinction.

66 million years ago, The Cretaceous. 76% of species lost.

Volcanic activity and climate change was the cause for the ammonites in the ocean as well there

was an asteroid impact that brought the dinosaur’s end during this time.

Now we want to look at

t=597 million years

This was when the earth went through a dramatic change and there was a big explosion of life

(The Cambrian). This is when the Moon predicted the second as exactly 1. So we have the

starting point for the great diversification of life, the Cambrian. Our time line becomes

597 million years ago, the Cambrian

444 million years ago, the Ordovician

375 million years ago, The Devonian

251 million years ago, The Permian

200 million years ago, the Triassic

66 million years ago, The Cretaceous

Let us plot these…

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We see the trend is 153 million years, 69 million years, 124 million years, 51 million years, 134

million years are the times from extinction event to extinction event. They alternate roughly 150,

50, 150, 50…

So, since from the Triassic the one next to the Cretaceous (dinosaurs) is 150, the next would be

50 million years after the Cretaceous plus or minus about 20 million years. The Cretaceous is at

66 million years ago so 50 plus 20 is 70 million years. We are at 66 million years since the

Cretaceous. So we are at the zone between 50 million years after the Cretaceous and 70 million

years after the Cretaceous meaning we could be, if we fit it in the graph, at extinction now plus

or minus several million years. However, we think we are in an extinction called the

Anthropocene (Man made extinction). But we can change that, we know what we need to do to

stop man-made climate change.

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Thus we have a theoretical value for the radius of the proton:

Equation 65.

And a theoretical value for its charge

Equation 66.

Equation 67.

And we have the radius of a proton in terms of the solar magnetic field at Earth

Equation 68.

All of this based on the idea that the basis of their structure is in six-fold unfolding. Given our

constant k

Equation 69.

And making the approximation we can with these equations eliminate in equation

4 using equation 2 in which we can eliminate (see Appendix 3 where I do this and verify the

result for both units and values and verify other equations as well that we derived earlier).

Equation 70.

This has an accuracy as shown in the appendix of close to 88% because in equation 2 the charge

of six protons is predicted by the theory to be 5.72 protons, the rest of the equations are much

more accurate, but we seek to rectify that. We write this equation so we can have an equation

that defines the solar magnetic field by solving for .

To address the accuracy of the equation for the charge of a proton, equation 2, we ask what is

the culprit. We suggest it is . So we solve the equation for that to see by how much it is

off. It is:

Equation 71.

We see it should be 7.79573E-11 and is actually 7.885E-11. The value we are using is 98.86785%

accurate, but we want to do better (See Appendix 3).

2π

⋅

𝔼

(

⋅

)

h 4π r

= q

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71pr oton s ≈ 6pr oton s

(

2GM

⊙

⋅

(IE

)

⊙

)

1/3

k =

8π

(1 + α) ⋅ N

𝔼

1 + α = 1

31850496r

⋅

𝔼)

h 4π

(IE

)

⊙

(α

)/36

= q

⋅

⋅ k

h 4π r

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We now eliminate on the left in equation 70 with equation 65 to find the B field of the Sun as

described by the Earth orbit as the ground state:

Equation 72.

Where N=31850496 is a perfect integer, that is has no values after the decimal. Equation 8 gives

a magnetic field strength of 4.73E-3 Teslas (See Appendix 4). Concerning equation 7

We want it to be more accurate. To do that we have to substitute for , the fraction

which is 39/5 and we have to formulate a theory for why this would be, though the discrepancy

could be in as its experimental value has the largest errors. Other possibilities are the

is right and that as well this is due to the radius of a proton having large errors, or even that it is

supposed to the factor 8, which would be good to consider because 8-fold symmetry is very

dynamic, in particular in its role with beryllium 8 being a precursor to carbon in nuclear

synthesis by stars.

Because we are looking at 6 protons we had , and because of six-fold symmetry we had

and because the charge we had because of the 1/3 root on the right. So taking

and leaving the that came from substituting for k, and the as

well, and leaving the factors 2 and 4 because they describe the physical dynamics of the equation

we have:

Equation 73.

Where

Or we can factor out all the numbers on the right the 2 and 4 in equation 73 and write

Equation 74.

In which case N=31850496/4=7962624. We see the solar magnetic field is determined by the

radius of a hydrogen atom, its ionization energy, the solar gravitational field, with the earth orbit

as the ground state.

⊙

⋅

(IE

)

⊙

⋅

h 4π

= q

⋅

⋅ k

h 4π r

(α

)/36

(α

)/36

q = 6 q

N = 6

= 279936

8 = 2

⊙

⋅

(IE

)

2GM

⊙

⋅

h 4π

N = 6

= 279936

⊙

⋅

(IE

)

⊙

⋅

h π

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Appendix 1

We have our equation for the Earth-Moon-Sun Orbital System (Equation 6 on page 4)

And our equation for the proton (Equations 4 and 5 on page 4)

Where is six protons is carbon (C). We also have that for the Earth orbital velocity

(Equation 16 on page 7)

This results in

Which gives

This is

Because k (Equation 18, page 7) is in s/m is 1/k=773.5 m/s. Thus we have something like a

quantum state on the left but for the planets equal to a quantum state for the proton on the

right, and conclude k connects the larger scale macrocosmos to the microcosmos of the atom. Or

perhaps better to write:

Because then we have

Because . It makes one think of how the energy of a wave is the amplitude

squared. Here amplitude then is in seconds and energy is the square of time.

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

α ⋅ 6m

h 4π r

= 1secon d

k v

≈ 6

K E

moon

K E

earth

⋅ Ear th Da y =

k v

h 4π r

(

K E

moon

K E

earth

Ear th Da y

)

⋅

h 4π r

k (m ⋅ s) = s

k v

(

K E

moon

K E

earth

Ear th Da y

)

⋅

h 4π r

= s

k v

⋅

= 1

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Is actually equal to 1.2 seconds seconds squared, but we can say 1.2 seconds is a rough sketch for

the idea that it is one second, which you can do because in any physical theory there is room for

play for the physics to still serve its function. So we will evaluate it at one second. We have

On the right we have

This is an accuracy of about 95%.

Where 6 days is 6 rotations of the Earth. Just how accurate is this?

For the factor in terms of a proton we have

Since the mass of the earth and the mass of the moon do not vary, and the velocity of the moon

and the earth vary a little from aphelion to perihelion, their orbits are a little eccentric, we will

focus on the range of velocities.

(

K E

moon

K E

earth

Ear th Da y

)

k v

= 38.5

⋅

h 4π r

= 36.555

h 4π r

⋅

= 6d ays

h 4π r

18769

1.67262E − 27kg

(6.62607E − 34Js)(4π)(0.833E − 15m)

(6.67408E − 11Nm

/kg

)(299,792,459m /s)

(1.12213E 31)(5.37368E − 31) = 6.0299675 ≈ 6.03 ≈ 6.0secon d s

(6.03s)

⋅

= 6d ays

= 5.972E 24kg

= 7.34767E 22kg

6d a ys = 6(24)(60)(60) = 518400secon d s

= 81.2775

(6.03s)(81.2775) = 490.103 ≈ 490secon d s

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The measured values are, which are mean orbital velocities of the earth and moon:

Thus the accuracy of the equation is 29/32.5=90%. But, using the velocity of the moon at

aphelion which is 0.966 km/s and the velocity of the earth at perihelion which is 30.29km/s, we

have 30.29/0.966=31.356. This gives us an accuracy for our equation of 31.356/32.5=96.5%.

If we take

And divide each side by , then on the left we get which is the rest energy of a proton. We

pick up then 6 days divided by the speed of light squared on the right and have equations 1 and 2

on page 3:

The rest energy of a proton is

Since kinetic energy is

We have

518400

490

= 1057.959 ≈ 1058

= 1058

= 32.5

= 29.78

k m

= 1.022

k m

29.78k m /s

1.022k m /s

= 29

h 4π r

⋅

= 6d ays

4π r

K E

≈

6d a ys

κ ⋅

Pr otonSur fac Area

Pr oton RestEnerg y

⋅

Ear thKin et icE nerg y

Moon KineticEnerg y

6d a ys

LightSpeedSqu ared

κ =

= kg ⋅

= 18,769

(6.626E − 34)

(6.674E − 11)(299792459)

= 3.411557E − 12kg ⋅

= m

= (1.672E − 27)(299,792,459)

= 1.5E − 10Joules

of 51 62

Or at perihelion

Or at aphelion

The radius of a proton is

The surface area of a proton is

We have

Where 6 days is

The accuracy of our equation is

Which is closer to 6 days than 5 days or 7 days and we round to the nearest integer to get 6 days.

= 5.972E 24kg

= 7.34767E 22kg

= 29,780m /s

= 30,290m /s

= 1022

= 966m /s

= 0.833E − 15m

4π r

= 8.72E − 30m

4π r

= 2.9529E − 15m

5.972E 24)(30,290)

7.34767E 22)(966)

= 79,912.451

6d a ys

= (3.41E − 12) ⋅

2.953E − 15

1.5E − 10

⋅ (79,912.451) = 5.3646E − 12

6d a ys /c

= 6(24)(60)(60)/(8.98755E16) = = 5.768E − 12

5.3646E − 12

5.768E − 12

100 = 93.0

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Appendix 2

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Appendix 3

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Appendix 4

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The Author